Rajiv, Sanju and Tanu started a business with the investment of Rs. (z-400), (z-1000) and (z+200) respectively. After four months, Rajiv decreased his investment by 50% and Sanju increased his investment by 30% and Tanu decreased his investment by Rs. ‘y’. At the end of one year, the profit share of Sanju will be 50% more than the profit share of Rajiv. At the end of one year, the profit share of Sanju and Tanu will be the same. Which of the following options represents the correct relation between the values of ‘y’ and ‘z’?

Solution

Ratio of Rajiv, Sanju and Tanu investment with respect to the time =(z-400) x4+(100-50) % of (z-400) x8 : (z-1000)x4+(100+30)% of (z-1000)x8 : (z+200)x4+(z+200-y)x8    Eq.(i) The profit share of Sanju will be 50% more than the profit share of Rajiv. profit share of Sanju = (100+50) % of profit share of Rajiv the profit share of Sanju = 150% of the profit share of Rajiv profit share of Sanju = 1.5 x (profit share of Rajiv) the profit share of Sanju: profit share of Rajiv = 1.5/1 = 15/10 = 3:2    Eq.(ii) At the end of one year, the profit share of Sanju and Tanu will be the same. profit share of Sanju: profit share of Tanu ⇒ 3:3    Eq.(iii) From Eq.(ii) and Eq.(iii), the profit share of Rajiv: profit share of Sanju: profit share of Tanu ⇒ 2:3:3    Eq.(iv) From Eq.(i) and Eq.(iv). [(z-400)x4+(100-50)% of (z-400)x8]/[(z-1000)x4+(100+30)% of (z-1000)x8] = 2/3 [(z-400)x4+50% of (z-400)x8]/[(z-1000)x4+130% of (z-1000)x8] = 2/3 [(z-400) x4+0.5(z-400) x8]/[(z-1000)x4+1.3(z-1000)x8] = ⅔ So,  z = 4000 Put the value of ‘z’ in Eq.(i) The ratio of Rajiv, Sanju, and Tanu investment with respect to the time ⇒ (4000-400) x4+(100-50) % of (4000-400) x8: (4000-1000)x4+(100+30)% of (4000-1000)x8 : (4000+200)x4+(4000+200-y)x8 ⇒ 3600x4+50% of 3600x8: 3000x4+130% of (3000) x8: (4200) x4+(4200-y) x8 ⇒ 3600+50% of 3600x2: 3000+130% of (3000) x2: (4200) +(4200-y) x2 ⇒ 3600+1800x2: 3000+3900x2: (4200) +(4200-y) x2 ⇒ 3600+3600: 3000+7800: (4200) +(8400-2y) ⇒ 7200: 10800: (12600-2y) …. Eq.(v) From Eq.(iii) and Eq.(v). 10800 / (12600-2y) = 3/3 10800 = (12600-2y) 10800 = (12600-2y) 2y = 12600-10800 = 1800 y = 9 00 a) 5y = (z+800) 5x900 = (4000+800) 4500 = 4800 The above equation is not satisfied. So, this is not the correct answer. b) (z+200)/7 = y (4000+200)/7 = 900 4200/7 = 600 600 = 900 The above equation is satisfied. So, this is not the correct answer. c) 6y = (z+200) 6x900 = (4000+200) 5400 = 4200 The above equation is not satisfied. So this is not the correct answer. d) (z+400)/6 = 1.5y (4000+400)/6 = 1.5x900 (4000+400)/6 = 900 4400 = 8100 The above equation is not satisfied. So, this is not the correct answer.